121. Best Time to Buy and Sell Stock #
题目 #
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Note that you cannot sell a stock before you buy one.
Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Not 7-1 = 6, as selling price needs to be larger than buying price.
Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
题目大意 #
给定一个数组，它的第 i 个元素是一支给定股票第 i 天的价格。如果你最多只允许完成一笔交易（即买入和卖出一支股票），设计一个算法来计算你所能获取的最大利润。注意你不能在买入股票前卖出股票。
解题思路 #
- 题目要求找出股票中能赚的钱最多的差价
- 这一题也有多个解法，可以用 DP，也可以用单调栈
代码 #
package leetcode
// 解法一 模拟 DP
func maxProfit(prices []int) int {
if len(prices) < 1 {
return 0
}
min, maxProfit := prices[0], 0
for i := 1; i < len(prices); i++ {
if prices[i]-min > maxProfit {
maxProfit = prices[i] - min
}
if prices[i] < min {
min = prices[i]
}
}
return maxProfit
}
// 解法二 单调栈
func maxProfit1(prices []int) int {
if len(prices) == 0 {
return 0
}
stack, res := []int{prices[0]}, 0
for i := 1; i < len(prices); i++ {
if prices[i] > stack[len(stack)-1] {
stack = append(stack, prices[i])
} else {
index := len(stack) - 1
for ; index >= 0; index-- {
if stack[index] < prices[i] {
break
}
}
stack = stack[:index+1]
stack = append(stack, prices[i])
}
res = max(res, stack[len(stack)-1]-stack[0])
}
return res
}